Answer
$$\frac{{12 - 3{\pi ^2}}}{{16}}$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /2} {{x^3}\cos 2x} dx \cr
& {\text{Using integration by parts method }} \cr
& \,\,\,\,\,{\text{Let }}u = {x^3},\,\,\,\,du = 3{x^2}dx,\,\,\,\,\,dv = \cos 2xdx,\,\,\,\,v = \frac{1}{2}\sin 2x \cr
& \cr
& {\text{Integration by parts then gives}} \cr
& \int {{x^3}\cos 2x} dx = {x^3}\left( {\frac{1}{2}\sin 2x} \right) - \int {\left( {\frac{1}{2}\sin 2x} \right)} \left( {3{x^2}dx} \right) \cr
& \int {{x^3}\cos 2x} dx = \frac{{{x^3}}}{2}\sin 2x - \int {\frac{3}{2}{x^2}\sin 2xdx} \cr
& \cr
& {\text{Integrate by parts again}} \cr
& \,\,\,\,\,{\text{Let }}u = \frac{3}{2}{x^2},\,\,\,\,du = 3xdx,\,\,\,\,\,dv = \sin 2xdx,\,\,\,\,v = - \frac{1}{2}\cos 2x \cr
& \int {{x^3}\cos 2x} dx = \frac{{{x^3}}}{2}\sin 2x - \left( { - \frac{3}{4}{x^2}\cos 2x - \int {\left( { - \frac{1}{2}\cos 2x} \right)\left( {3xdx} \right)} } \right) \cr
& \int {{x^3}\cos 2x} dx = \frac{{{x^3}}}{2}\sin 2x + \frac{3}{4}{x^2}\cos 2x - \int {\frac{3}{2}x\cos 2x} dx \cr
& \cr
& {\text{Integrate by parts again}} \cr
& \,\,\,\,\,{\text{Let }}u = \frac{3}{2}x,\,\,\,\,du = \frac{3}{2}dx,\,\,\,\,\,dv = \cos 2xdx,\,\,\,\,v = \frac{1}{2}\sin 2x \cr
& \int {{x^3}\cos 2x} dx = \frac{{{x^3}}}{2}\sin 2x + \frac{3}{4}{x^2}\cos 2x - \left( {\frac{3}{4}x\sin 2x - \int {\left( {\frac{1}{2}\sin 2x} \right)\left( {\frac{3}{2}dx} \right)} } \right) \cr
& \int {{x^3}\cos 2x} dx = \frac{{{x^3}}}{2}\sin 2x + \frac{3}{4}{x^2}\cos 2x - \frac{3}{4}x\sin 2x + \frac{3}{4}\int {\sin 2x} \cr
& \int {{x^3}\cos 2x} dx = \frac{{{x^3}}}{2}\sin 2x + \frac{3}{4}{x^2}\cos 2x - \frac{3}{4}x\sin 2x - \frac{3}{8}\cos 2x + C \cr
& {\text{Then}} \cr
& \int_0^{\pi /2} {{x^3}\cos 2x} dx = \left( {\frac{{{x^3}}}{2}\sin 2x + \frac{3}{4}{x^2}\cos 2x - \frac{3}{4}x\sin 2x - \frac{3}{8}\cos 2x} \right)_0^{\pi /2} \cr
& = \left( {\frac{{{{\left( {\pi /2} \right)}^3}}}{2}\sin \left( \pi \right) + \frac{3}{4}{{\left( {\frac{\pi }{2}} \right)}^2}\cos \left( \pi \right) - \frac{3}{4}\left( {\frac{\pi }{2}} \right)\sin \left( \pi \right) - \frac{3}{8}\cos \left( \pi \right)} \right) - \left( { - \frac{3}{8}\cos 0} \right) \cr
& = \frac{{{\pi ^3}}}{{16}}\left( 0 \right) + \frac{3}{{16}}{\pi ^2}\left( { - 1} \right) - \frac{1}{4}\left( {\frac{\pi }{2}} \right)\left( 0 \right) - \frac{3}{8}\left( { - 1} \right) + \frac{3}{8} \cr
& = - \frac{3}{{16}}{\pi ^2} + \frac{3}{4} \cr
& = \frac{{12 - 3{\pi ^2}}}{{16}} \cr} $$
